Question

căn bậc 3 (x-3) + 3 căn(2x+1)=10

Answer 1

ĐKXĐ: x>=-1/2

\(\sqrt[3]{x-3}+3\sqrt{2x+1}=10\)

=>\(\sqrt[3]{x-3}-1+3\sqrt{2x+1}-9=0\)

=>\(\dfrac{x-3-1}{\sqrt[3]{\left(x-3\right)^2}+\sqrt[3]{x-3}+1}+3\left(\sqrt{2x+1}-3\right)=0\)

=>\(\dfrac{x-4}{\sqrt[3]{\left(x-3\right)^2}+\sqrt[3]{x-3}+1}+3\cdot\dfrac{2x+1-9}{\sqrt{2x+1}+3}=0\)

=>\(\left(x-4\right)\left(\dfrac{1}{\sqrt[3]{\left(x-3\right)^2}+\sqrt[3]{x-3}+1}+\dfrac{6}{\sqrt{2x+1}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)